# Is gravity the Earth's centripetal acceleration? Should the gravitational acceleration be equal to the centripetal acceleration at the equator?

I understand that different forces can act as centripetal forces (shear, tension of a string etc) but in the case of the rotating earth, is it really the gravitational force the centripetal force that keeps the earth spinning? In that case, shouldn't the centripetal acceleration at the equator be equal to the gravitational acceleration of 9.81 m/s2, instead of the 0.03 m/s2 obtained using the linear velocity at the equator and the equatorial radius? What force actually acts as the centripetal force on earth or on any rotating sphere?

I have also read elsewhere that for someone standing on earth, the centripetal force is the net force resulting from the difference between the weight and the normal force. But where does this difference comes from (why is it different from 0)? As the gravitational attraction itself doesn't change, I believe this difference is due to a change in the normal force. But why does it change at all?

First, no force is needed on a spinning object to keep it spinning. Second, force is needed to cause objects to move on non-straight paths. Third, don't try to classify an single force as a centripetal force.

When you analyze the motion of an object, identify all the actual forces (as vectors) acting on the object, ideally using a free-body diagram. Then find the net force by adding those vectors. If the object is moving in a circular path, you then set the radial component of that net force equal to the mass, $$m$$, of the object times the radial, or centripetal, acceleration term: $$\left(\sum_i \vec{F}_i\right)\cdot\hat{r} = m \vec{a}_c = m\frac{v^2}{r}= m\omega^2 r.$$

In the case of an object at the surface of the earth, there is no single force which is a "centripetal force." There are forces which contribute to the centripetal acceleration. Those forces are the gravitational force the earth's mass exerts on the object mass and the normal component of the contact (electromagnetically-based) force between the object's bottom surface and the surface the object is resting on.

Mathematically you have $$\frac{v^2}{R_e}=\frac{mg - F_{norm}}{m}$$

I wrote it that way to emphasize that the centripetal acceleration is the sum of forces in the radial direction divided by the mass. That's a much better way to think about these concepts rather than trying to find "the centripetal force."

P.S.- Don't make the mistake of thinking that the normal component of contact force is equal to mg. It's not.

• I wish I could give this answer eight up-votes, one for each point of clarification of a commonly misunderstood concept. – garyp Feb 17 '20 at 2:20
• @garyp Thanks. That's very encouraging.This is what I try to do in my introductory classes. – Bill N Feb 17 '20 at 16:57

Gravity in the case of spinning celestial bodies can be compared to the tensile strenght of a cable attached to a spinning weight.

The centripetal force is always: $$F_c = m\omega^2r$$. If the cable will break or not due to that force is another subject.

If a spinning asteroid has an acceleration of gravity too small compared to the required centripetal acceleration, it can't hold any loose rock (not to mention air and water).